/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 The rear window defogger of a ca... [FREE SOLUTION] | 91Ó°ÊÓ

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The rear window defogger of a car consists of thirteen thin wires (resistivity \(=88.0 \times 10^{-8} \Omega \cdot \mathrm{m}\) ) embedded in the glass. The wires are connected in parallel to the \(12.0-\mathrm{V}\) battery, and each has a length of 1.30 \(\mathrm{m} .\) The defogger can melt \(2.10 \times 10^{-2} \mathrm{kg}\) of ice at \(0^{\circ} \mathrm{C}\) into water at \(0^{\circ} \mathrm{C}\) in two minutes. Assume that all the power delivered to the wires is used immediately to melt the ice. Find the cross-sectional area of each wire.

Short Answer

Expert verified
The cross-sectional area of each wire is approximately \(3.57 \times 10^{-6} \text{ m}^2\).

Step by step solution

01

Calculate Energy Needed to Melt the Ice

We start by calculating the energy needed to melt the given mass of ice. The formula for the energy required to melt ice is \( Q = mL_f \), where \( m \) is the mass of ice and \( L_f \) is the latent heat of fusion for ice (\(334 \text{ kJ/kg}\)). Substitute \( m = 2.10 \times 10^{-2} \text{ kg} \):\[ Q = 2.10 \times 10^{-2} \text{ kg} \times 334 \times 10^3 \text{ J/kg} = 7014 \text{ J} \]
02

Calculate the Power Needed

The wires melt the ice in two minutes, so we convert this time to seconds: \( t = 120 \text{ s} \). Power is the rate at which energy is used, given by \( P = \frac{Q}{t} \).Substituting the values, we have:\[ P = \frac{7014 \text{ J}}{120 \text{ s}} = 58.45 \text{ W} \]
03

Find the Total Current Through the Wires

Since the total power delivered is 58.45 W, the total current \( I \) drawn by the wires can be calculated using the formula \( P = IV \), where \( V = 12.0 \text{ V} \).Rearranging the formula gives:\[ I = \frac{P}{V} = \frac{58.45 \text{ W}}{12.0 \text{ V}} = 4.87 \text{ A} \]
04

Calculate the Resistance of One Wire

Because the wires are in parallel, each wire carries an equal part of the current. The current through each wire is \( I_{wire} = \frac{I}{n} \), where \( n = 13 \).\[ I_{wire} = \frac{4.87 \text{ A}}{13} = 0.374 \text{ A} \]
05

Calculate the Resistance in One Wire Using Ohm's Law

For one wire, the potential difference is also \( 12.0 \text{ V} \). Using Ohm's law, \( V = IR \), we find the resistance \( R \) of one wire:\[ R = \frac{V}{I_{wire}} = \frac{12.0 \text{ V}}{0.374 \text{ A}} = 32.09 \text{ \(\Omega\)} \]
06

Calculate Cross-sectional Area Using Resistance Formula

With the resistance known, use the formula \( R = \frac{\rho L}{A} \), where \( \rho = 88.0 \times 10^{-8} \Omega \cdot \mathrm{m} \) is the resistivity and \( L = 1.30 \text{ m} \) is the length of the wire, to solve for cross-sectional area \( A \):Rearranging gives [using Algebra]:\[ A = \frac{\rho L}{R} = \frac{88.0 \times 10^{-8} \Omega \cdot \mathrm{m} \times 1.30 \text{ m}}{32.09 \Omega} \]
07

Perform the Calculation for Cross-sectional Area

Now calculate the area:\[ A = \frac{88.0 \times 10^{-8} \times 1.30}{32.09} \approx 3.57 \times 10^{-6} \text{ m}^2 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistivity and Its Calculation
Resistivity is a fundamental property of materials that measures how strongly they oppose the flow of electric current. It's typically denoted by the Greek letter \( \rho \), and its units are ohm-meters (\( \Omega \cdot \text{m} \)). To calculate resistivity, you need to understand the resistance of the material along with its dimensions. In the case of the car's rear window defogger, we calculated each wire's resistance using the resistivity formula: - The formula is given by \( R = \frac{\rho L}{A} \), where: - \( R \) is the resistance. - \( \rho \) is the resistivity. - \( L \) is the length of the wire. - \( A \) is the cross-sectional area.In our exercise, with an already known resistance \( R \), the resistivity \( \rho \), and the length \( L \), we solve for \( A \). This calculation illustrates how important understanding resistivity is for designing efficient electrical systems.
Understanding Ohm's Law
Ohm's Law is a critical principle in the study of electricity. It relates the voltage (\( V \)), current (\( I \)), and resistance (\( R \)) in a simple formula: \( V = IR \). This foundational law helps us understand how current flows through a conductor when voltage is applied. For the car defogger system:- We needed to calculate the resistance of each wire.- Using the voltage supplied by the battery and the current flowing through each wire, we found this resistance.- Ohm's Law helps ensure that electrical components like the defogger wires operate safely and effectively by keeping the voltage, current, and resistance in balance.Grasping Ohm's Law allows for the correct sizing and selection of electrical components in circuits.
Latent Heat of Fusion
The latent heat of fusion is the energy required to change a substance from solid to liquid at its melting point, without altering its temperature. For water, this is about 334 kJ/kg. It's a vital concept in thermal physics since it quantifies the energy needed to melt ice in our context:- To calculate how much energy was needed to melt the ice off the car's rear window, we used the formula \( Q = mL_f \). - \( Q \) is the heat energy required. - \( m \) is the mass of the ice. - \( L_f \) is the latent heat of fusion for water.Understanding latent heat ensures that we can accurately compute the energy transfers in thermal systems.
Power and Energy Relationship
The relationship between power and energy is crucial in electric circuits. Power (\( P \)) is defined as the rate at which energy (\( Q \)) is used or transferred, expressed as \( P = \frac{Q}{t} \), where \( t \) is the time in seconds. In practical electrical systems, like our defogger wires:- You need to calculate the energy used by the system.- Determine how efficiently this energy is transformed into work, such as melting ice in this case.- With known energy consumption and time, you can determine the power output.This relation helps evaluate the effectiveness of an electrical system, ensuring it provides the necessary energy for its intended task without excessive energy loss.

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Most popular questions from this chapter

When a "dry-cell" flashlight battery with an internal resistance of 0.33\(\Omega\) is connected to a \(1.50-\Omega\) light bulb, the bulb shines dimly. However, when a lead-acid "wet-cell" battery with an internal resistance of 0.050\(\Omega\) is connected, the bulb is noticeably brighter. Both batteries have the same emf. Find the ratio \(P_{\text { wet }} / P_{\text { dry }}\) of the power delivered to the bulb by the wet-cell battery to the power delivered by the dry-cell battery.

A battery has an internal resistance of 0.50\(\Omega .\) A number of identical light bulbs, each with a resistance of \(15 \Omega,\) are connected in parallel across the battery terminals. The terminal voltage of the battery is observed to be one-half the emf of the battery. How many bulbs are connected?

A piece of Nichrome wire has a radius of \(6.5 \times 10^{-4} \mathrm{m} .\) It is used in a laboratory to make a heater that uses \(4.00 \times 10^{2} \mathrm{W}\) of power when connected to a voltage source of 120 \(\mathrm{V}\) . Ignoring the effect of temperature on resistance, estimate the necessary length of wire.

A battery delivering a current of 55.0 \(\mathrm{A}\) to a circuit has a terminal voltage of 23.4 \(\mathrm{V}\) . The electric power being dissipated by the internal resistance of the battery is 34.0 \(\mathrm{W}\) . Find the emf of the battery.

The coil of a galvanometer has a resistance of \(20.0 \Omega,\) and its meter deflects full scale when a current of 6.20 \(\mathrm{mA}\) passes through it. To make the galvanometer into a nondigital ammeter, a 24.8 \(\mathrm{m} \Omega\) shunt resistor is added to it. What is the maximum current that this ammeter can read?

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